Til the Cows Come Home
Time Limit: 1000MS   Memory Limit: 65536K
Total Submissions: 38556   Accepted: 13104

Description

Bessie is out in the field and wants to get back to the barn to get as much sleep as possible before Farmer John wakes her for the morning milking. Bessie needs her beauty sleep, so she wants to get back as quickly as possible.

Farmer John's field has N (2 <= N <= 1000) landmarks in it, uniquely numbered 1..N. Landmark 1 is the barn; the apple tree grove in which Bessie stands all day is landmark N. Cows travel in the field using T (1 <= T <= 2000) bidirectional cow-trails of various lengths between the landmarks. Bessie is not confident of her navigation ability, so she always stays on a trail from its start to its end once she starts it.

Given the trails between the landmarks, determine the minimum distance Bessie must walk to get back to the barn. It is guaranteed that some such route exists.

Input

* Line 1: Two integers: T and N

* Lines 2..T+1: Each line describes a trail as three space-separated integers. The first two integers are the landmarks between which the trail travels. The third integer is the length of the trail, range 1..100.

Output

* Line 1: A single integer, the minimum distance that Bessie must travel to get from landmark N to landmark 1.

Sample Input

5 5
1 2 20
2 3 30
3 4 20
4 5 20
1 5 100

Sample Output

90
注意:先输入边数后输入结点数,存在重边
#include"cstdio"
using namespace std;
const int MAXN=;
const int INF=0x3fffffff;
int mp[MAXN][MAXN];
int V,E;
int vis[MAXN];
int d[MAXN];
int dijkstra(int s)
{
for(int i=;i<=V;i++)
{
vis[i]=;
d[i]=mp[s][i];
}
vis[s]=; for(int i=;i<=V;i++)
{
int mincost,k;
mincost=INF;
for(int j=;j<=V;j++)
{
if(!vis[j]&&d[j]<mincost)
{
k=j;
mincost=d[j];
}
} vis[k]=;
for(int j=;j<=V;j++)
{
if(!vis[j]&&d[j]>d[k]+mp[k][j])
{
d[j]=d[k]+mp[k][j];
}
} }
return d[];
}
int main()
{
while(scanf("%d%d",&E,&V)!=EOF)
{
for(int i=;i<=V;i++)
for(int j=;j<=V;j++)
if(i==j) mp[i][j]=;
else mp[i][j]=INF;
for(int i=;i<E;i++)
{
int u,v,cost;
scanf("%d%d%d",&u,&v,&cost);
if(cost<mp[u][v]) mp[u][v]=mp[v][u]=cost;//存在重边
}
int ans=dijkstra(V);
printf("%d\n",ans);
}
return ;
}

堆优化的dijkstra

#include"cstdio"
#include"vector"
#include"queue"
using namespace std;
typedef pair<int,int> P;
const int MAXN=;
const int INF=0x3fffffff;
int mp[MAXN][MAXN];
int V,E;
vector<int> G[MAXN];
int d[MAXN];
void dijkstra(int s,int end)
{
for(int i=;i<=V;i++) d[i]=INF; priority_queue<P, vector<P>,greater<P> > que;
que.push(P(,s));
d[s]=; while(!que.empty())
{
P p=que.top();que.pop();
if(p.second==end)
{
printf("%d\n",p.first);
return ;
}
int v=p.second;
if(d[v]<p.first) continue;
for(int i=;i<G[v].size();i++)
{
int to=G[v][i];
if(d[to]>d[v]+mp[v][to])
{
d[to]=d[v]+mp[v][to];
que.push(P(d[to],to));
}
}
}
}
int main()
{
while(scanf("%d%d",&E,&V)!=EOF)
{
for(int i=;i<=V;i++)
{
G[i].clear();
for(int j=;j<=V;j++)
if(i==j) mp[i][j]=;
else mp[i][j]=INF;
}
for(int i=;i<E;i++)
{
int u,v,cost;
scanf("%d%d%d",&u,&v,&cost);
G[v].push_back(u);
G[u].push_back(v);
if(cost<mp[u][v]) mp[v][u]=mp[u][v]=cost;
}
dijkstra(,V);
}
return ;
}

spfa+前向星可解决重边问题。

#include <cstdio>
#include <cstring>
#include <queue>
using namespace std;
const int MAXN=;
const int INF=0x3f3f3f3f;
struct Edge{
int to,w,next;
}es[];
int head[MAXN],tot;
int n,m;
void addedge(int u,int v,int w)
{
es[tot].to=v;
es[tot].w=w;
es[tot].next=head[u];
head[u]=tot++;
}
int d[MAXN],vis[MAXN];
void spfa(int s)
{
for(int i=;i<=n;i++)
{
d[i]=INF;
vis[i]=;
}
queue<int> que;
que.push(s);
d[s]=;
vis[s]=;
while(!que.empty())
{
int u=que.front();que.pop();
vis[u]=;
for(int i=head[u];i!=-;i=es[i].next)
{
Edge e=es[i];
if(d[e.to]>d[u]+e.w)
{
d[e.to]=d[u]+e.w;
if(!vis[e.to])
{
que.push(e.to);
vis[e.to]=;
}
}
}
}
printf("%d\n",d[n]);
}
int main()
{
while(scanf("%d%d",&m,&n)!=EOF)
{
memset(head,-,sizeof(head));
tot=;
for(int i=;i<m;i++)
{
int u,v,w;
scanf("%d%d%d",&u,&v,&w);
addedge(u,v,w);
addedge(v,u,w);
}
spfa();
}
return ;
}

Java版:

前向星+spfa

import java.util.Arrays;
import java.util.LinkedList;
import java.util.Scanner;
import java.util.Queue;
class Edge{
int to,w,net;
Edge(){}
Edge(int to,int w,int net)
{
this.to=to;
this.w=w;
this.net=net;
}
}
public class Main{
static final int MAXN=1005;
static final int INF=0x3f3f3f3f;
static int m,n;
static int[] head = new int[MAXN];
static Edge[] es = new Edge[4005];
static int tot;
static void addedge(int u,int v,int w)
{
es[tot] = new Edge(v,w,head[u]);
head[u] = tot++;
} static int[] d = new int[MAXN];
static boolean[] vis = new boolean[MAXN];
static int spfa(int src,int ter)
{
Arrays.fill(vis, false);
Arrays.fill(d, INF);
Queue<Integer> que = new LinkedList<Integer>();
que.add(src);
d[src]=0;
while(!que.isEmpty())
{
int u=que.peek();que.poll();
vis[u]=false;
for(int i=head[u];i!=-1;i=es[i].net)
{
Edge e = es[i];
if(d[e.to]>d[u]+e.w)
{
d[e.to]=d[u]+e.w;
if(!vis[e.to])
{
que.add(e.to);
vis[e.to]=true;
}
}
}
}
return d[ter];
}
public static void main(String[] args){
Scanner in = new Scanner(System.in);
while(in.hasNext())
{
tot=0;
Arrays.fill(head, -1);
m=in.nextInt();
n=in.nextInt();
for(int i=0;i<m;i++)
{
int u,v,w;
u=in.nextInt();
v=in.nextInt();
w=in.nextInt();
addedge(u,v,w);
addedge(v,u,w);
}
int res=spfa(n,1);
System.out.println(res);
}
}
}

POJ2387(最短路入门)的更多相关文章

  1. 图论:HDU2544-最短路(最全、最经典的最短路入门及小结)

    最短路 Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submis ...

  2. POJ - 2387 Til the Cows Come Home (最短路入门)

    Bessie is out in the field and wants to get back to the barn to get as much sleep as possible before ...

  3. poj2387 最短路

    题意:给出一堆双向路,求从N点到1点的最短路径,最裸的最短路径,建完边之后直接跑dij或者spfa就行 dij: #include<stdio.h> #include<string. ...

  4. POJ1502(最短路入门题)

    MPI Maelstrom Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 7471   Accepted: 4550 Des ...

  5. [原]最短路专题【基础篇】(updating...)

    hud1548 a strange lift  最短路/bfs  题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1548 题意:一个奇怪的电梯,每层楼的 ...

  6. 【最短路】Dijkstra+ 链式前向星+ 堆优化(优先队列)

    Dijkstra+ 链式前向星+ 优先队列   Dijkstra算法 Dijkstra最短路算法,个人理解其本质就是一种广度优先搜索.先将所有点的最短距离Dis[ ]都刷新成∞(涂成黑色),然后从起点 ...

  7. POJ2387 Til the Cows Come Home (最短路 dijkstra)

    AC代码 POJ2387 Til the Cows Come Home Bessie is out in the field and wants to get back to the barn to ...

  8. POJ-2387(原始dijkstra求最短路)

    Til the Cows Come Home POJ-2387 这题是最简单的最短路求解题,主要就是使用dijkstra算法,时间复杂度是\(O(n^2)\). 需要注意的是,一定要看清楚题目的输入要 ...

  9. poj2387 初涉最短路

    前两天自学了一点点最短路..看起来很简单的样子... 就去kuangbin的专题找了最简单的一道题练手..然后被自己萌萌的三重for循环超时虐的不要不要的~ 松弛虽然会但是用的十分之不熟练... 代码 ...

随机推荐

  1. TOML简介 (转)

    TOML的由来 配置文件的使用由来已久,从.ini.XML.JSON.YAML再到TOML,语言的表达能力越来越强,同时书写便捷性也在不断提升. TOML是前GitHub CEO, Tom Prest ...

  2. mongodb的mongod.lock文件及oplog文件

    在mongodb的启动时,在数据目录下,会生成一个mongod.lock文件.如果在正常退出时,会清除这个mongod.lock文件,若要是异常退出,在下次启动的时候,会禁止启动,从而保留一份干净的一 ...

  3. excel表格定义导入到powerdesigner脚本

    打开powerdesigner,shift + ctrl + X 打开脚本窗口 输入执行的脚本,点 run 即可. 简单的导入Excel脚本 '开始 Option Explicit Dim mdl ' ...

  4. Oracle 10g ORA-12154: TNS: could not resolve the connect identifier specified 问题解决! 我同事遇到的问题。 username/

    Oracle 10g ORA-12154: TNS: could not resolve the connect identifier specified 问题解决! 我同事遇到的问题. userna ...

  5. UnicodeEncodeError: ‘ascii’ codec can’t encode characters in position xxx ordinal not in range(12

    python在安装时,默认的编码是ascii,当程序中出现非ascii编码时,python的处理常常会报这样的错UnicodeDecodeError: 'ascii' codec can't deco ...

  6. Unix环境高级编程—进程控制(三)

    一.解释器文件 解释器文件属于文本文件,起始行形式为: #! pathname[optional-argument] 我们创建一个只有一行的文件如下: #!/home/webber/test/echo ...

  7. 【BZOJ4826】[Hnoi2017]影魔 单调栈+扫描线

    [BZOJ4826][Hnoi2017]影魔 Description 影魔,奈文摩尔,据说有着一个诗人的灵魂.事实上,他吞噬的诗人灵魂早已成千上万.千百年来,他收集了各式各样的灵魂,包括诗人.牧师.帝 ...

  8. JAVA中int与String类型的相互转换

    Java的int和String类型间互相转换,小功能但是经常用到,下面是几种实现的方法: 字符串类型String转换成整数int 1. int i = Integer.parseInt([String ...

  9. phpPHP创建创建jpg格式图片以及压缩图片(转)

    其实是因为一些业务上的需求,所以需要对用户上传后的图片进行压缩,因为上传的图片比较大,显示的时候加载起来如果网速不给力的话就很吃力了,而且大图片也浪费空间,于是找了一下相关的资源,主要方法在开源中国上 ...

  10. CUDA: 常量内存与事件

    常量内存: 常量内存用于保存在核函数执行期间不会发生变化的数据,在变量面前添加  __constant__  修饰符: __constant__  Sphere  s[SPHERES]; cudaMe ...